Question
Download Solution PDFIn PLL as frequency synthesizer, if the crystal oscillator having frequency 500 MHz is passed through divide by 2 network, then the input frequency to PLL is _______ MHz.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In a PLL (Phase Locked Loop) used as a frequency synthesizer, a crystal oscillator provides a highly stable frequency. If this frequency is passed through a divide-by-N network, the output frequency becomes \( \frac{f}{N} \).
Given:
Crystal oscillator frequency, \( f = 500~\text{MHz} \)
Divide-by-2 network: N =2
Calculation:
Input frequency to PLL = \( \frac{500}{2} = 250~\text{MHz} \)
Final Answer:
2) 250 MHz
Last updated on Jun 7, 2025
-> RRB JE CBT 2 answer key 2025 for June 4 exam has been released at the official website.
-> Check Your Marks via RRB JE CBT 2 Rank Calculator 2025
-> RRB JE CBT 2 admit card 2025 has been released.
-> RRB JE CBT 2 city intimation slip 2025 for June 4 exam has been released at the official website.
-> RRB JE CBT 2 Cancelled Shift Exam 2025 will be conducted on June 4, 2025 in offline mode.
-> RRB JE CBT 2 Exam Analysis 2025 is Out, Candidates analysis their exam according to Shift 1 and 2 Questions and Answers.
-> The RRB JE Notification 2024 was released for 7951 vacancies for various posts of Junior Engineer, Depot Material Superintendent, Chemical & Metallurgical Assistant, Chemical Supervisor (Research) and Metallurgical Supervisor (Research).
-> The selection process includes CBT 1, CBT 2, and Document Verification & Medical Test.
-> The candidates who will be selected will get an approximate salary range between Rs. 13,500 to Rs. 38,425.
-> Attempt RRB JE Free Current Affairs Mock Test here
-> Enhance your preparation with the RRB JE Previous Year Papers.